Question

Sn=1/1.4+1/2.5+1/3.6+........+tn find its value and also find find S∞

Answer 1

of course it is

Answer 2

are those decimals underneath or multiplication?

Answer 3

it's multiplication sign

Answer 4

\[\frac{1}{n+\frac{n+3}{10}}\]
\[\frac{10}{11n+3}\]
\[\frac{10}{3(\frac{11}{3}n+1)}\]
\[\frac{10}{3}\sum_1 \frac{1}{\frac{11}{3}n+1}=\frac{10}{3}\sum_1 \frac{1}{kn+1}x^{kn+1}\]

Answer 5

.... oh, decimal would be more interesting lol

Answer 6

\[\sum_1^N\frac{1}{n(n+3)}\]

Answer 7

decompose that to get:\[\frac{1}{3n}-\frac{1}{3(n+3)}\]

Answer 8

\[\frac13-\frac{1}{12}\]\[\frac16-\frac{1}{15}\]\[\frac19-\frac{1}{18}\]\[\frac1{12}-\frac{1}{21}\]\[\frac1{15}-\frac{1}{24}\]
its telescoping now

Answer 9

1/3 + 1/6 + 1/9 - (1/3(n+3) + 1/3(n+2)+1/3(n+1))

Answer 10

its not decimal it's in multiplication'

Answer 11

a corrected for that :)

Answer 12

but yur ans is not cuming i also got it till here only...

Answer 13

\[\frac13+\frac16+\frac19-\left(\frac{1}{3(n+3)}+\frac{1}{3(n+2)}+\frac{1}{3(n+1)}\right)\]

Answer 14

simplify that to your hearts content for tn

Answer 15

when n=infty, the right side drops to zero

Answer 16

\[tn=\frac{ 1 }{ n \left( n+3 \right) }=\frac{ 1 }{ n \left( 0+3 \right) }+\frac{ 1 }{-3\left( n+3 \right) }\]
\[tn=\frac{ 1 }{ 3n }-\frac{ 1 }{ 3\left( n+3 \right) }\]
\[tn=\frac{ 1 }{ 3 }\left( \frac{1 }{ n }- \frac{ 1 }{n+3 }\right)\]
substitute n=1,2,3,4,.............,n-3,n-2,n-1,n and adding vertically downwards

Answer 17

it will give us 1/3(1-1/4)
1/3(1/2-1/5)

Answer 18

i would contend that we start at n=4 for mine

Answer 19

n=1 is fine .. just had some invalid concern :)

Answer 20

for n = 100 000 000, this gets closer and closer to the first 3 terms

http://www.wolframalpha.com/input/?i=1%2F3%2B1%2F6%2B1%2F9+-+%281%2F%283%28n%2B3%29%29%2B1%2F%283%28n%2B2%29%29%2B1%2F%283%28n%2B1%29%29%29%2C+n%3D100000000

Answer 21

but what will be the result of it

Answer 22

what do u think the ans will be